# Help with quantum physics

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Hello guys, i need some help in clarifying a question.

A beam of ultraviolet light with wavelength of 200nm is incident on a metal whose work function is 3.0eV. Note that this metal is applied with +1.0V with respect to the ground, Determine the largest kinetic energy of the photoelectrons generated in this process.

Using the formula of KEmax = hf - work function, i can come up with an answer for it. However, from the question, it states the metal is applied with +1.0V with respect to the ground;Does this Voltage affect the KEmax? Because if i were to find stopping potential Vs using E=q|Vs|, Vs will result in approximately 6.211V, taking q to be 1.6 x 10^-19. So i believe that this voltage across the metal is redundant in the calculation.

Thank you guys!

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DrClaude
Mentor
Using the formula of KEmax = hf - work function, i can come up with an answer for it. However, from the question, it states the metal is applied with +1.0V with respect to the ground;Does this Voltage affect the KEmax? Because if i were to find stopping potential Vs using E=q|Vs|, Vs will result in approximately 6.211V, taking q to be 1.6 x 10^-19.
So if the applied potential was 6.211 V instead of 1 V, what would you answer for KEmax?

xTyler
Won't it be the same as KEmax?

I'm quite confused, is it alright to assume that the applied +1V on the metal is also the stopping voltage? And does this applied voltage on the metal makes it harder for the electrons to break the bond from metal -> Vacuum -> Being ejected ?

DrClaude
Mentor
I'm quite confused, is it alright to assume that the applied +1V on the metal is also the stopping voltage?
No. As said, the stopping potential is 6.211 V.

And does this applied voltage on the metal makes it harder for the electrons to break the bond from metal -> Vacuum -> Being ejected ?
No. It changes what happens to the freed electron.

DrClaude
Mentor
Won't it be the same as KEmax?
What is the definition of the stopping potential?

I had this question figured out. Thanks!